poj 2533 Longest Ordered Subsequence

不发题目了
其实就是输入n个数，然后求最长上升子序列的长度（严格上升）
之前写的二分出现小问题，不停RE
最后怒用STL
有一个很清晰的讲解 ​​传送门​​

---

【代码】

//poj 2533 Longest Ordered Subsequence
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define fo(i,j,k) for(i=j;i<=k;i++)
using namespace std;
int n,len=1;
int a[10001],d[10001];
int main()
{
    int i,j,k;
    scanf("%d",&n);
    fo(i,1,n)
      scanf("%d",&a[i]);
    d[1]=a[1];
    fo(i,2,n)
    {
        if(a[i]>d[len]) d[++len]=a[i];
        else if(a[i]<d[1]) d[1]=a[i];
        else d[lower_bound(d,d+len,a[i])-d]=a[i];
    }
    printf("%d\n",len);
    return 0;
}

---

【还有一个手写二分代码】

//poj 2533 Longest Ordered Subsequence
#include<iostream>
#include<cstdio>
#include<cstring>
#define fo(i,j,k) for(i=j;i<=k;i++)
using namespace std;
int n,len=1;
int a[10001],d[10001];
inline int erfen(int x)
{
    int l=1,r=len;
    while(l<=r)
    {
        int mid=(l+r)/2;
        if(d[mid-1]<x && x<=d[mid]) return mid;
        if(x>d[mid]) l=mid+1;
        else r=mid-1;
    }
}
int main()
{
    int i,j,k;
    scanf("%d",&n);
    fo(i,1,n)
      scanf("%d",&a[i]);
    d[1]=a[1];
    fo(i,2,n)
    {
        if(a[i]>d[len]) d[++len]=a[i];
        else if(a[i]<d[1]) d[1]=a[i];
        else d[erfen(a[i])]=a[i];
    }
    printf("%d\n",len);
    return 0;
}

注意 if(d[mid-1]